Lanjutan Analisa Hubung Singkat
Transcript of Lanjutan Analisa Hubung Singkat
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Short
Circuit
nalysis
o
0.10 020
030
..... t
0.40 0.50
Fig. 6.5 Electrical torque
on
three phase termlnaJ short circuit.
Field current after
short circuit
o -_+-_-_-_-_
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ .;
t ----
Fig. 6.6 Oscilogram of the field current after a short circuit.
6.10 Effect of Load Current or Prefault Current
185
Consider a 3-phase synchronous generator supplying a balanced 3-phase load. Let a three
phase fault occur at the load terminals. Before the fault occurs, a load current
IL
is flowing into
the load from the generator. Let the voltage at the fault be v and the terminal voltage of the
generator be
Vt
Under fault conditions, the generator reactance
is
xd
The circuit
in
Fig. 6.7 indicates the simulation of fault at the load terminals by a parallel
switch
S
E; =
V
t
+ j Xd
IL
= VI + Xext + j
xd)I
L
where E;
is
the subtransient internal voltage.
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Power System Analysis
s
Fault
Fig. 6.7
For the transient state
~
=
V j
xd
L
= V
f
Zext j xd )IL
E
or
~ are used only when there
is
a prefault current I
L
. Otherwise E
g
the steady
state voltage in series with the direct axis synchronous reactance
is
to be used for all calculations.
Eg
remains the same for all L values, and depends only on the field current. Every time, of
course, a new E
is
required to be computed.
6 11 Reactors
Whenever faults occur in power system large currents flow. Especially,
if
the fault is a dead
short circuit at the terminals
or
bus bars t?normous currents flow damaging the equipment and
its components. To limit the flow
of
large currents under there circumstances current limiting
reactors are used. These reactors are large coils covered for high self-inductance.
They are also so located that the effect of the fault does not affect other parts of the
system and is thus localized. From time to time new generating units are added to an existing
system to augment the capacity. When this happens, the fault current level increases and it
may become necessary to change the switch gear. With proper use of reactors addition of
generating units does not necessitate changes in existing switch gear.
6 12 Construction
of
Reactors
These reactors are built with non magnetic core so that saturation
of
core with consequent
reduction in inductance and increased short circuit currents is avoided. Alternatively, it is
possible to use iron core with air-gaps included in the magnetic core so that saturation is avoided.
6 13 Classification
of
Reactors
i) Generator reactors, ii) Feeder reactors,
iii) Bus-bar reactors
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Circuit
nalysis
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The above classification is based on the location
ofthe
reactors. Reactors may be connected in
series with the generator in series with each feeder or to the bus bars.
i) Generator reactors
The reactors are located in series with each of the generators as shown in
Fig. 6.8 so that current flowing into a fault F from the generator
is
limited.
Generators
Bus
Bars
Fig. 6.8
Disadvantages
(a)
In
the event
of
a fault occuring on a feeder, the voltage at the remaining
healthy feeders also
may
loose synchronism requiring resynchronization later.
(b) There is a constant voltage drop in the reactors and also power loss, even
during normal operation. Since modern generators are designed to with stand
dead short circuit at their terminals, generator reactors are now-a-days not
used except for old units
in
operation.
(ii) Feeder reactors In this method of protection each f ~ e e r is equipped with a
series reactor as shown
in
Fig 6.9.
In the event of a fault on any feeder the fault current drawn is restricted by the
reactor.
Generators
~ ~ r ~ ~ ~ ~ B V S
Reactors
Bars
JC ~
Fig. 6.9
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F
Power System Analysis
Disadvantages:
I. Voltage drop and power loss still occurs
in
the reactor for a
feeder fault. However, the voltage drop occurs only in that particular feeder reactor.
2
Feeder reactors do not offer any protection for bus bar faults. Neverthless,
bus-bar faults occur very rarely.
As series reactors inhererbly create voltage drop, system voltage regulation will
be impaired. Hence they are to be used only
in
special case such as for short
feeders
of
large cross-section.
iii) Bus bar reactors:
In both the above methods, the reactors carry full load current
under normal operation. The consequent disadvantage of constant voltage drops
and power loss can be avoided by dividing the bus bars into sections and inter
connect the sections through protective reactors. There are two ways
of
doing
this.
a)
Ring system :
In
this method each feeder is fed by one generator. Very little power flows
across the reactors during normal operation. Hence, the voltage drop and
power loss are negligible. If a fault occurs on any feeder, only the generator
to which the feeder
is
connected will feed the fault and
o ~ h r
generators are
required to feed the fault through the reactor.
b)
Tie bar system:
This
is
an improvement over the ring system. This
is
shown
in Fig. 6.11. Current fed into a fault has to pass through two reactors in
series between sections.
F F
BVS
Bars
BVS
bar
Tie
bar
Generators
F
Peeders
Fig.
6.10
Fig. 6.11
Another advantage
is
that additional generation may be connected to the
system without requiring changes
in
the existing reactors.
The only disadvantage
is
that this systems requires an additional bus-bar
system, the tie-bar.
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nalysis
189
Worked Examples
E
6.1
Two generators rated
at 10 MVA, 11
KV and 15
MVA, 11
KV respectively
are
connected in parallel to a bus. The bus bars feed two motors rated 7.5 MVA and
10 MVA respectively. The rated voltage of the motors is 9 KV. The reactance of
each generator is
12
and that of each
motor
is
15
on
their
own ratings.
Assume 50 MVA, 10 KV base and draw the reactance diagram.
Solution
The reactances of the generators and motors are calculated on 50 MVA, 10 KV base values.
Reactance
of
generator I = X
G
=
12
. Cr ~ ~ ) = 72.6
(
11)2 (50)
Reactance of generator 2 = X
G2
= 12
1 . 1
= 48.4
Reactance of motor I = X
M1
= 15 .
r
=
81
Reactance of motor 2 = X
M2
= 15 (1
9
0
r J
= 60.75
The reactance diagram is drawn and shown in Fig. E.6.1.
Fig. E.6.1
E.6.2 A 100 MVA, 13.8
KV,
3-phase generator has a reactance
of
20 . The generator
is
connected to a 3-phase transformer T rated 100 MVA 12.5 KV 1110 KV with 10
reactance. The h.v. side of the transformer is connected to a transmission line of
reactance 100 ohm. The
far
end
of
the line
is
connected to a step down transformer
T
2
made of three single-phase transformers each rated 30
MVA,
60 KV /
10 KV
with 10 reactance the generator supplies two motors connected on the
l.v.
side
T2
as shown in Fig. E.6.2. The motors are rated
at
25
MVA
and 50
MVA
both at
10 KV
with
15
reactance. Draw the reactance diagram showing all the values
in
per
unit. Take generator rating as base.
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ower System Analysis
Solution
Base MVA 100
Base KV 13.8
110
Base KV for the line = 13.8 x 12.5 = 121.44
,J3
x66KV 114.31
Line-to-line voltage ratio ofT2
10KV
= w
121.44xl0
,
Base voltage for motors 114.31 10.62
KV
X for generators 20 0.2
p.ll.
(
12.5J2 100
X for transformer
T
10
x 13.8 x 100 8.2
X for transformer T2 on
,J3
x 66 : 10 KV and 3 x 30 MVA base 10
X for T2 on 100 MVA, and 121.44 KV: 10.62 KV
is
X T2
= 10
x
C ~ ~ 2 r
x C ~ ~ J
=
9.85 0.0985 p.ll.
(
121.44J2
Base reactance for line
1
147.47 ohms
100
Reactance of line 147.47 0.678
p.ll.
(
10 J2 90J
Reactance of motor M
lOx
10.62 25 31.92
0.3192 p.ll.
(
10 J2 90J
Reactance of motor
M2
= 10 x 10.62 50 = 15.96
The reactance diagram is shown in Fig. E.6.2.
Fig E S 2
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Circuit nalysis 191
E.6.3 Obtain the per unit representation for the three-phase power system shown in
Fig.
E.6.3.
Generator 1 : 50 MVA
Generator
2 :
25 MVA
Generator 3 : 35 MVA.,
Transformer
T1 : 30 MVA,
Transformer T2 : 25 MVA
Fig. E.6.3
10.5
KV;
X = 1 8 ohm
6.6 KV; X
=
1 2
ohm
6.6 KV; X
=
0.6 ohm
11/66 KV, X
=
15 ohm/phase
66/6.2
KV,
as h.v. side X = 12 ohms
Transmission line : XL = 20 ohm/phase
Solution
Let base MVA = 50
base KV = 66 L- L)
Base voltage on transmission as line 1
p.u.
(66 KV)
Base voltage for generator I :
11
KV
Base voltage for generators 2 and 3 : 6.1 KV
20x50
p.u. reactance of transmission line = 66
2
= 0.229 p.u.
1 5
x50
p.ll.
reactance
of
transformer T] = = 0.172
p.u.
12x 50
p.ll.
reactance
of
transformer
T2
=
6T
=
0.1377
p.u.
l. x
50
p.u.
reactance
of
generator 1 =
(11)2
= 0.7438
p.u.
1 2
x 50
p.u.
reactance
of
generator 2
=
(6.2)2
=
1.56
p.u.
0.6x50
p.u. reactance of generator 3 = (6.2)2 = 0.78 p.u.
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92 Power System Analysis
E 6.4 A single phase two winding
transformer
is rated 20 KVA, 480/120 V at 50 HZ. The
equivalent leakage impedance of
the
transformer
referred to
I.v.
side is 0.0525
78.13 ohm using
transformer
ratings as base values,
determine
the
per
unit
leakage impedance referred to the h.v. side and
t.v.
side.
olution
Let base KVA = 20
Base voltage on h.v. side
=
480 V
Base voltage on l.v. side
=
120 V
The leakage impedance on the l.v. s i ~ of the transformer
= Z = V
base
2
2
VA
base
120)2
20,000 = 0.72 ohm
p.LI. leakage impedance referred
to
the l.v. of the transformer
0.0525 78.13
= p u
2
= = 0.0729 78.13
0.72
Equivalent impedance referred to h.v. side is
r 0.0525 70.13] = 0.84 78.13
480)2
The base impedance on the h. v. side of the transformer is 20,000 = 11.52 ohm
p.LI.
leakage impedance referred to
h.v.
side
0.84 78.13 = 0.0729 78.13
p. u.
11.52
E.6.5
A
single phase
transformer is
rated
at
110/440
V
3
KVA.
Its leakage reactance
measured on
110
V side
is
0.05 ohm. Determine the leakage impedance referred
to 440 V side.
Solution
0.11)2 x 1000
Base impedance on
110
V side
=
3
=
4.033 ohm
0.05
Per unit reactance on 110 V side = 4.033 = 0.01239 p.u.
440)2
Leakage reactance referred to 440 V side = 0.05) 11 = 0.8 ohm
. 0.8
Base impedance referred to 440 V side
=
64.53
=
0.01239 p.u.
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Short Circuit nalysis
193
E.6.6 Consider the system shown in Fig. E.6.4. Selecting 10 000 KVA and 11
Vas
base values find the p.u. impedance ofthe 200 ohm load referred to 11 KV side
and 11 KV side.
II KV
1I0KV
IIOKV 55KV
~ o o o V A ~ Q ~ ~ ~
200 o m
X =9
X
I
~ 7 : ; '
1
Fig E 6 4
Solution
Base voltage at p = KY
110
Base voltage at R = ~ = 55 K Y
Base impedance at R =
55
2
x 1000
10,000
200 ohm
= 302.5 ohm
p.u. impedance at R = 302.5 ohm = 0.661 ohm
Base impedance at
=
110
2
x 1000
10,000
= 1210 ohm
Load impedance referred to = 200 x 22 = 800 ohm
800
p.lI. impedance
of
load referred to = 1210 = 0.661
Similarly base impedance at P =
112 x 1000
10,000
= 121.1 ohm
Impedance
of
load referred to P = 200 x 22 x 0.12 = 8 ohm
8
p.lI. impedance
of
load at P = 12 1 = 0.661 ohm
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Power System Analysis
E.6.7 Three
transformers each rated
30 MVA
at
38.1/3.81 KV
are connected
in
star-delta
with a balanced load of three 0.5 ohm, star connected resistors. Selecting
a base of 900 MVA 66 KV for the h.v. side of the transformer find the base values
for
the
I
v.
side.
Solution
Fig E G 5
base
KY
L
_d
2
3.81)2
Base impedance on
I Y
side
= =
0.1613 ohm
Base
MYA
90
0.5
p.lI. load resistance on 1 Y side = 0.1613 = 3.099 p.lI.
Base impedance on
h.Y.
side
=
66)2
=
48.4 ohm
90
66 J2
Load resistance referred to h.Y. side
=
0.5 x 3.81 = 150 ohm
150
p.lI. load resistance referred to
h.Y.
side
=
48.4
=
3.099 p.lI.
The per unit load resistance remains the same.
E.6.8 Two
generators
are connected in parallel to
the I.v.
side of a 3-phase
delta-star
transformer
as
shown
in Fig. E.6.6.
Generator
1 is
rated
60,000 KVA
11
KV.
Generator 2 is rated 30,000 KVA 11
KV.
Each generator
has
a subtransient
reactance
of
xd =
25 .
The transformer is rated 90,000 KVA at
11
KV f / 66 KV
with a reactance
of
10 . Before a fault occurred the voltage on the h.t. side
of
the transformer
is 63 KY. The transformer
in
unloaded
and there
is no circulat ing
current between the generators. Find the
subtransient
current in each
generator
when a
3-phase
short circuit
occurrs
on the h.t. side of the
transformer.
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Short Circuit nalysis
60 000 KVA
KV
'-'1-----,
KV 66 KV
Solution:
'-' r - - - -
30,OOOKVA
KV
Fig E S S
l1 /Y
Let the line voltage on the h.v. side be the base KV = 66 KV.
Let the base
KVA
= 90 000
KVA
t
90 000
Generator 1 : xd = 0.25 x 60 000 = 0.375 p.u.
90 000
For generator 2 :
xd =
30 000
=
0.75 p.u.
The internal voltage for generator I
0.63
Eg1
=
0.66
=
0.955 p.u.
The internal voltage for generator
2
0.63
Eg2 = 0.66 = 0.955 p.u.
195
The reactance diagram is shown
in
Fig. E.6.7 when switch S is closed the fault condition
is
simulated. As there
is no
circulating current between the generators the equivalent reactance
0.375 x 0.75
of
the parallel circuit
is
0.375 0.75
=
0.25 p.u.
j
0.10
s
Fig E S 7
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196
Power
System Analysis
Th b
0 .955 . 27285
e Sll traQslent current I = =
J
p.lI.
(j0.25
jO.1
0)
The voltage as the delta side
of
the transformer is
-j
2.7285)
G
0.10)
=
0.27205 p.u.
I;
=
the subtransient current flowing into fault from generator
0.955 - 0.2785
I; =
j 0.375
1.819 p.u.
0.955 - 0.27285
Similarly,
2 =
j 0.75
=
-j 1.819 p.u.
The actual fault currents supplied
in
amperes are
1.819 x 90,000
II = r =
8592.78 A
,3 x II
0.909
x
90,000
I
=
J3
x II = 4294.37 A
E.6.9 R station with two generators feeds through transformers a transmission system
operating
at
132 KV. The far end of the transmission system consisting of 200 km
long double circuit line is connected to load from bus B. f a 3-phase fault occurs
at
bus
B,
determine the total fault current and fault current supplied by each
generator.
Select 75
MVA
and
I1
KVon LV side and 132 KVon
h.v.
side as base values.
111132 KV
75 MVA A
75MVA
B
200Km
10%
25MVA
~ 5 ~
0.189 ohm/phaselKm
10% 8%
32 KV
Fig. E.S.S
Solution
p.
u. x
of generator 1 = j 0.15 p.
u.
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Short ircuit
nalysis
75
p.u. x of generator 2 =
j
= 0.10 25
=
j
0.3 p.u.
p.u. x of transformer T, =
j 0 1
75
p.u. x
of
transformer T2 = j 0.08 x 25 = j 0.24
j 0 180x200x75 .
p u xofeachline= 132xl32 =JO 1549
The equivalent reactance diagram is shown in Fig. E.6.9 (a), (b) (c).
j
0.5
j
0 1
j
0.1549
jO.1549
a)
j 0.7745
jO.17 + jO.07745 = j 0.2483
b)
c)
Fig. E.6.9
Fig. E.6.9 (a), (b) (c) can be reduced further into
Zeq = j 0.17 + j 0.07745 = j 0.248336
I
o
Total fault current = - J 4.0268 p
u
j 0.248336
75 x 1000
Base current for 132 KY circuit = J x 132 = 328 A
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Power
System
nalysis
Hence actual fault current = - j 4.0268 x 328 = 32 A
L 9
0
75 x 1000
Base current for KV side of the transformer = {;; = 3936.6 A
, ,xlI
Actual fault current supplied from KV side
=
3936.6
x
4.0248 ; = 1585 J 9 A L 9
0
585 39
L - 90 x j 0.54 .
Fault current supplied
by
generator I
=
j 0.54 j 0.25
=
-J 10835.476 A
15851.9xj
0.25
Fault current supplied by generator 2 = j 0.79 = 5016.424 A
L 90o
E.6.10 A 33 KV line has a resistance of 4 ohm and reactance of
6
ohm respectively.
The line is connected to a generating station bus bars through a 6000 KVA step
up
transformer which has a reactance
of
6%.
The station has two generators
rated 10,000 KVA with 10 reactance and 5000 KVAwith 5 reactance. Calculate
the fault current antl short circuit KVA when a 3-phase fault occurs at the h.v.
terminals of the transformers and at the load end of the line.
Solution:
10,000 KVA
10
rv}---. .
v} __ l
5
5,000 KVA
60,000 KVA
~ ,
I
Fig. E.S.10 a)
33 KV
4
6
Let 10,000 KVA be the base KVA
Reactance of generator 1 Xo 1 = 10
5 x 10,000
Reactance of generator 2 X
02
=
5000
=
10
6 x 10,000
Reactance of transformer X
T
= 6 000 = 10
,
The line impedance is converted into percentage impedance
KVA. X 10,000x16
X = ; X = = 14.69
10(KVf Lme IOx(33)2
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Short
ircuit nalysis
19000x 4
R
Line
= =
3.672
10(33)2
199
(i) For a 3-phase fault at the h.y. side terminals
of
the transformer fault impedance
(
IOXI0)
=
+ 10=
15
10 + 10
Fig 6.10 (b)
10,000 x 100
Short circuit
KVA
fed into the fault = 15 KYA
=
66666.67 KYA
=
66.67
MYA
For a fault at F2 the load end
of
the line the total reactance to the fault
= 15 +
l4.69
=
29.69
Total resistance to fault = 3.672
Total impedance to fault =
~ 3 6 7 2 2
+ 29.69
2
= 29.916
100
Short circuit KVA into fault = 29.916 x 10,000
=
33433.63 KYA
=
33.433 MYA
E.6.11
Figure E.6.11 (a) shows a power system where load at bus 5
is
fed by generators
at bus 1 and bus 4. The generators are rated at 100 MVA;
11
KV with subtransient
reactance of 25 . The transformers are rated each at 100 MVA
11 112
KVand
have a leakage reactance of 8 . The lines have an inductance of 1 mH phase
km. Line
L1
is 100
krri
long while lines
L2
and L
J
are
each of 50 km in length.
Find the fault current and
MVA
for a 3-phase fault at bus S.